Swift错误：无法找到接受提供参数的'&&'的重载

我一直在为Swift编写一些教程。我遇到了一个TicTacToe教程，我尝试使用Xcode 6 Beta 6进行编码。当我检查字典中的值时，出现以下错误：无法找到接受所提供参数的'& &'的过载。这是我的代码。Swift错误：无法找到接受提供参数的'&&'的重载

var plays = [Int:Int]() 

var whoWon = ["I":0,"you":1] 
for (key,value) in whoWon { 
if ((plays[6] == value && plays[7] == value && plays[8] == value) || 
    (plays[3] == value && plays[4] == value && plays[5] == value) || 
    (plays[0] == value && plays[1] == value && plays[2] == value) || 
    (plays[6] == value && plays[3] == value && plays[0] == value) || 
    (plays[7] == value && plays[4] == value && plays[1] == value) || 
    (plays[8] == value && plays[5] == value && plays[2] == value) || 
    (plays[6] == value && plays[4] == value && plays[2] == value) || // error appears on this line 
    (plays[8] == value && plays[4] == value && plays[0] == value)) 
{ 
    userMessage.hidden = false 
    userMessage.text = "Looks like \(key) won!" 
}

来源

2014-09-03 Renee

如何将某些东西分解为子表达式？ – skyguy 2014-09-20 16:14:09

如果你看一下在报告导航完整编译器的输出，那么你将看到消息

note: expression was too complex to be solved in reasonable time; consider breaking up the expression into distinct sub-expressions

它告诉你如何解决这个问题。

来源

2014-09-03 21:47:34

谢谢。我忘了问题导航器选项卡。 – Renee 2014-09-03 22:17:00

嘿@Martin R我正在使用这个相同的教程，我是一个初学者，你会如何将它分解成子表达式？ – skyguy 2014-09-20 02:46:38

我在做同样的教程。在我看来有点不可思议，它在子表达式被打破了，但它的伎俩，当我读到这里： http://swiftlang.eu/community/34-xcode-6-beta-6-swift-can-t-handle-long-expressions/0

这可能是由于Xcode中的问题。

这里的重写功能：

func checkForWin() { 
    var whoWon = ["I": 0, "You": 1] 
    for (key, value) in whoWon { 
     var wonA = (plays[1] == value && plays[2] == value && plays[3] == value) 
     var wonB = (plays[4] == value && plays[5] == value && plays[6] == value) 
     var wonC = (plays[7] == value && plays[8] == value && plays[9] == value) 
     var wonD = (plays[1] == value && plays[4] == value && plays[7] == value) 
     var wonE = (plays[2] == value && plays[5] == value && plays[8] == value) 
     var wonF = (plays[3] == value && plays[6] == value && plays[9] == value) 
     var wonG = (plays[1] == value && plays[5] == value && plays[9] == value) 
     var wonH = (plays[3] == value && plays[5] == value && plays[7] == value) 

     if(wonA || wonB || wonC || wonD || wonE || wonF || wonG || wonH) { 
       userMessage.hidden = false 
       userMessage.text = "Looks like \(key) won!" 
       resetBtn.hidden = false 
       done = true 
     } 
    } 
}

来源

2014-10-14 13:38:07

为什么'var'而不是'let'？（I.E.，won *可能是常量，对吗？...） – 2014-10-14 13:54:09

使用圆括号“分手表达成不同的子表达式”。 This works in Xcode 6.4

if (((plays[1] == value) && (plays[2] == value) && (plays[3] == value)) || 
    ((plays[4] == value) && (plays[5] == value) && (plays[6] == value)) || 
    ((plays[7] == value) && (plays[8] == value) && (plays[9] == value)) || 
    ((plays[1] == value) && (plays[4] == value) && (plays[7] == value)) || 
    ((plays[2] == value) && (plays[5] == value) && (plays[8] == value)) || 
    ((plays[3] == value) && (plays[6] == value) && (plays[9] == value)) || 
    ((plays[1] == value) && (plays[5] == value) && (plays[9] == value)) || 
    ((plays[3] == value) && (plays[5] == value) && (plays[7] == value))) {...}

来源

2015-07-11 07:45:07 Jessica

Swift错误：无法找到接受提供参数的'&&'的重载

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